User:Egm4313.s12.team3.landers

=Problem 1=

Problem Statement
For a spring-dashpot system in parallel with mass and an applied force f(t), derive the equation of motion. Figure of a spring-dashpot system in parallel:

Solution
The kinetics for this problem:
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$$\displaystyle y=y_k=y_c $$     (1)
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$$\displaystyle m\ddot{y}+y_I=f(t) $$     (2) $$y_I=$$ forces opposing f(t)
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$$\displaystyle y_I=y_k+y_c $$     (3) For a spring:
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$$\displaystyle y_k=kx_k $$     (4) For a dashpot:
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$$\displaystyle y_c=c\dot{x}_c $$     (5) Taking the derivative of equation 1, you recieve:
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$$\displaystyle \dot{y}=\dot{y_k}=\dot{y_c} $$     (6) Plug equations 4 and 5 into equation 3:
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$$\displaystyle y_I=kx_k+c\dot{x_c} $$     (7) Taking equation 7 and plugging it back into equation 2 yields:
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$$ f(t)=m\ddot{x}+c\dot{x_c}+kx_k $$
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Problem Statement
Find and plot the solution to the L2 ODE with no excitation.
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$$\displaystyle {y}''-10{y}'+25y=r(x) $$     (1) Given initial conditions:
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$$\displaystyle y(0)=1 $$
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$$\displaystyle {y}'(0)=0 $$
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 * due to there being no excitation:
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$$\displaystyle r(x)=0 $$
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Problem Solution
The Characteristic Equation
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$$\displaystyle \lambda^2+a\lambda+b=r(x)=0 $$     (2) From the original equation (1):
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$$\displaystyle a=-10 $$
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$$\displaystyle b=25 $$ Which yields:
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$$\displaystyle \lambda^2-10\lambda+25=0 $$    (3) Equation 3 can be simplified to:
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$$\displaystyle (\lambda-5)^2 $$ It has a double root;
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$$\displaystyle \lambda=5 $$ Also can be shown by discriminant:
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$$\displaystyle (-10)^2-(4)(25)=0 $$ The General Solution
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$$\displaystyle y=(c_1+c_2x)e^{5x} $$     (4) Take the Derivative in order to apply initial conditions
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$$\displaystyle {y}'=c_2e^{5x}+5(c_1+c_2x)e^{5x} $$    (5) Apply initial conditions into equations 4 and 5:
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$$\displaystyle y(0)=1=c_1 $$    (6)
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$$\displaystyle {y}'(0)=0=c_2+5c_1 $$    (7) Solving equations 6 and 7 yields:
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$$\displaystyle c_1=1 $$    (8)
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$$\displaystyle c_2=-5 $$    (9) Plugging the these values back into the general solution (4) yields:
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$$ y=(1-5x)e^{5x} $$
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